@wschrabi : what slopes do you want?  You can define a piecewise-linear function appropriately.  For example,

K := t -> piecewise(t <= 0, 0, t <.001, 5800*t/.001, t < 0.0104167 - .001, 5800, t < .0104167, 
(.0104167 - t)*5800/.001, 0);

@wschrabi : what slopes do you want?  You can define a piecewise-linear function appropriately.  For example,

K := t -> piecewise(t <= 0, 0, t <.001, 5800*t/.001, t < 0.0104167 - .001, 5800, t < .0104167, 
(.0104167 - t)*5800/.001, 0);

Right now the page <http://www.mapleprimes.com/recent/unread> is showing me a bunch of old postings and claiming they are recent, e.g.

where the last posting was from January 1 2009.

@otherworld314 : more information, including Maple's estimate of the error, is provided if you set infolevel[evalf] to at least 2 before doing the integration.

Since the y integral has a non-constant endpoint, Maple can't use specialized multiple-integration routines and must process this as an iterated integral.  However, it can be reformulated as a double integral over a rectangle, which works much faster:

> f := sqrt(1+cos(x)*cos(y)*cos(x+y));
   ft := eval(f, y = (x+1)*t);
   infolevel[evalf]:= 2;
   ti:= time():
   evalf(Int(Int(ft*(x+1),t=0..1),x=0..1));
   time()-ti;;

Control_multi:   integrating on [0, 0] .. [1, 1] the integrand

                                                    1/2
      (1 + cos(x) cos((x + 1) t) cos(x + (x + 1) t))    (x + 1)

trying DCUHRE (TOMS algorithm 698)
cuhre:   epsrel=.5000000000e-9; minpts=0; maxpts=195000
cuhre:   result=1.67430257749842592
cuhre:   abserr=.573230504794849768e-9; usedpts=1235
result=1.67430257749842592

                             1.674302577


                                  0.

@otherworld314 : more information, including Maple's estimate of the error, is provided if you set infolevel[evalf] to at least 2 before doing the integration.

Since the y integral has a non-constant endpoint, Maple can't use specialized multiple-integration routines and must process this as an iterated integral.  However, it can be reformulated as a double integral over a rectangle, which works much faster:

> f := sqrt(1+cos(x)*cos(y)*cos(x+y));
   ft := eval(f, y = (x+1)*t);
   infolevel[evalf]:= 2;
   ti:= time():
   evalf(Int(Int(ft*(x+1),t=0..1),x=0..1));
   time()-ti;;

Control_multi:   integrating on [0, 0] .. [1, 1] the integrand

                                                    1/2
      (1 + cos(x) cos((x + 1) t) cos(x + (x + 1) t))    (x + 1)

trying DCUHRE (TOMS algorithm 698)
cuhre:   epsrel=.5000000000e-9; minpts=0; maxpts=195000
cuhre:   result=1.67430257749842592
cuhre:   abserr=.573230504794849768e-9; usedpts=1235
result=1.67430257749842592

                             1.674302577


                                  0.

Gabriel's solution is basically the way to go, but you probably want to use the option arrows=none in at least two of your DEplot commands, to avoid having three direction fields interfering with each other.

Gabriel's solution is basically the way to go, but you probably want to use the option arrows=none in at least two of your DEplot commands, to avoid having three direction fields interfering with each other.

I think the question you had in mind was rather different from what you asked.  It seems you just want to define a piecewise function K.  That can be done as follows:

> K := t -> piecewise(t <= 0, 0, t < 0.0104167,5800,0);

I think the question you had in mind was rather different from what you asked.  It seems you just want to define a piecewise function K.  That can be done as follows:

> K := t -> piecewise(t <= 0, 0, t < 0.0104167,5800,0);

@otherworld314 : for the integral wrt y

> eval(Int(F(x,y),y=0..1),cos(x*a)=v);
   value(%) assuming a>0,a<Pi,v>0,v<1;


The result is rather messy, but seems correct.  For example:

> evalf(eval(%%=%,{a=2.3, v=0.7}));

                                                      -8
           1.245062228 = 1.245062228 - 0.1389670404 10   I

@otherworld314 : for the integral wrt y

> eval(Int(F(x,y),y=0..1),cos(x*a)=v);
   value(%) assuming a>0,a<Pi,v>0,v<1;


The result is rather messy, but seems correct.  For example:

> evalf(eval(%%=%,{a=2.3, v=0.7}));

                                                      -8
           1.245062228 = 1.245062228 - 0.1389670404 10   I

The symbolic version of pdsolve is quite limited when it comes to solving equations with BC's.  It can sometimes do separation-of-variables solutions, obtaining solutions from which (in some cases) you can construct a series solution.  For example:

> pdsolve({pde, f(0,t)=0, D[1](f)(1,t)=0});

works, but

> pdsolve({pde, f(0,t) = 0, D[1](f)(1,t) = 1});

doesn't (even though there is an easy solution, namely f(x,t)=x).

The symbolic version of pdsolve is quite limited when it comes to solving equations with BC's.  It can sometimes do separation-of-variables solutions, obtaining solutions from which (in some cases) you can construct a series solution.  For example:

> pdsolve({pde, f(0,t)=0, D[1](f)(1,t)=0});

works, but

> pdsolve({pde, f(0,t) = 0, D[1](f)(1,t) = 1});

doesn't (even though there is an easy solution, namely f(x,t)=x).

The bug seems to be that FunctionAdvisor has the sum_form for Psi as

Psi(t) = Sum((-1)^_k2*Zeta(_k2+2)*t^(_k2+1),_k2 = 0 .. infinity) for |t| < 1

where it should be -1/t - gamma + Sum((-1)^_k2*Zeta(_k2+2)*t^(_k2+1),_k2 = 0 .. infinity)

for 0 < |t| < 1.

In what context?  What kind of domains, and what is causing the walls to move?  If you can provide the appropriate equations, Maple can probably simulate the solutions.